Question

Can the mean value theorem for integrals be applied to an integral with an average value of 0?

Answer 1

Why not?

Answer 2

The requirement is a continuous function.

Answer 3

Okay, perhaps I did the problem wrong, but I followed the formula and got two number not in between the interval.

Answer 4

You'd have something like this:\[
f(c) = \frac{1}{b-a}\int_a^bf(x)~dx\to 0=\frac{1}{b-a}\cdot 0
\]

Answer 5

Okay, what is the problem?

Answer 6

f(x) = x3 − 9x on the interval [−1, 1]

Answer 7

x^3-9x from (-1,1)

Answer 8

$$\int_{-1}^{1} x^3-9x = f(c)(1-(-1)) \\

0 = f(c) * 2 \\
0= f(c)\\
0 = c^3 - 9c\\
0 = c(c^2 - 9)\\
c = 0 \text{ or} ~ c = 3, -3\\
\text {but 3 and -3 are not in the interval (-1,1)}\\
c = 0 \\
\text{0 is in the interval (-1,1)}


$$

Answer 9

Yeah, so in this case, \(f(0) = 0\), meaning \(c=0\).

Answer 10

Note that for any odd function f(-x) = -f(x)
$$ \int_{-a}^{a} f(x) =0$$

Answer 11

Okay, I see. Thank you very much.